Let be a polynomial function of degree 3 satisfying and . If product of the roots of the equation is 4 and the sum of all possible values of is then find . [Note: denotes greatest integer function less than or equal to .]

Road-map (why each step is chosen)

1. Translate the given equal values into an algebraic form. Because $f(0)=f(1)=f(2)=1$, it is natural to subtract 1 to create a cubic that vanishes at 0, 1 and 2.
2. Write the general cubic: $f(x)=a\,x(x-1)(x-2)+1$ (only one free parameter $a$ remains).
3. Use Vieta to relate roots’ product to $a$. For a cubic $ax^3+bx^2+cx+d=0$, we have product $= -\dfrac{d}{a}$. Here $d=f(0)=1$ and the product is given as 4, so $a$ is determined.
4. Evaluate $f(3)$ with that $a$. This is direct substitution.
5. Apply Greatest Integer Function. Finally round down and report the result.

Detailed walk-through

1. Creating a simpler cubic:

$f(0)=1,\; f(1)=1,\; f(2)=1 \implies f(x)-1 \text{ has roots } 0,1,2$

Thus

$f(x)-1=a\,x(x-1)(x-2)\;\;\Rightarrow\;\; f(x)=a\,x(x-1)(x-2)+1$

2. Relating $a$ to the product of roots of $f(x)=0$.
   The equation is

$a\,x(x-1)(x-2)+1=0$

Re-write in standard form: $ax^3-3a x^2+2a x+1=0$. Here

$d=1,\; a=\text{leading coefficient}$

So (Vieta)

$\text{product of roots}= -\frac{d}{a}=4 \implies a=-\frac14$

3. Compute $f(3)$.

$f(3)=a\,(3)(2)(1)+1=a\,6+1=-\frac14\times6+1=-\frac32+1=-\frac12$

4. Greatest Integer Function.

$\bigl[ f(3) \bigr]=\left[ -\frac12 \right]=-1$

What is the question about?

We have a special cubic (degree-3) polynomial. We know its value at three easy points (0, 1 and 2). From that, we want to describe the whole polynomial.

Why do we care about its roots?

The question tells us that when the polynomial is set equal to zero, the product of its three roots is 4. For cubics, Vieta’s formula connects the product of roots with the constant term and the leading coefficient.

What do we finally want?

After fixing the only possible polynomial that matches all the conditions, we plug in 3, find the value, and then apply the Greatest Integer Function (rounding down). The answer happens to be −1.

Let the cubic be $f(x)$ and let $f(0)=f(1)=f(2)=1$ (given).

1. Write in factor form

$f(x)=a\,x(x-1)(x-2)+1\quad(\text{for some }a\ne0)$

2. Standard form (needed for Vieta):

$f(x)=a\,x^3-3a\,x^2+2a\,x+1$

3. For $f(x)=0$ the product of roots is

$-\frac{d}{a}=-\frac{1}{a}$

Given this equals 4, so

$-\frac{1}{a}=4\;\;\Rightarrow\;\;a=-\frac14$

4. Evaluate $f(3)$:

$f(3)=a\,(3)(2)(1)+1=6a+1=6\left(-\frac14\right)+1=-\frac32+1=-\frac12$

5. Apply GIF:

$\bigl[f(3)\bigr]=\left[-\frac12\right]=-1$

Hence the required value is –1.